Question

$\mathrm{If}\sin 2A=\mathrm{\lambda }\sin 2B,\mathrm{prove}\mathrm{that}\frac{\tan (A+B)}{\tan (A-B)}=\frac{\mathrm{\lambda }+1}{\mathrm{\lambda }-1}$.

Answer 1

Given:
sin 2A = λ sin 2B

$\Rightarrow \frac{\sin 2A}{\sin 2B}=\lambda$

$$\begin{gathered} \Rightarrow \frac{\sin 2A+\sin 2B}{\sin 2A-\sin 2B}=\frac{\lambda +1}{\lambda -1} \\ \Rightarrow \frac{2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)}{2\sin \left(\frac{2A-2B}{2}\right)\cos \left(\frac{2A+2B}{2}\right)}=\frac{\lambda +1}{\lambda -1}\left[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\mathrm{and}\sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right] \\ \Rightarrow \frac{\sin \left(A+B\right)\cos \left(A-B\right)}{\sin \left(A-B\right)\cos \left(A+B\right)}=\frac{\lambda +1}{\lambda -1} \\ \Rightarrow \tan \left(A+B\right)\mathrm{co}t\left(A-B\right)=\frac{\lambda +1}{\lambda -1} \end{gathered}$$

$\Rightarrow \frac{\tan \left(A+B\right)}{\tan \left(A-B\right)}=\frac{\lambda +1}{\lambda -1}$

Hence proved.